Posted On February 19, 2014 |

Euclidean and coordinate geometry are both subjects you’ll want to be familiar with on the SAT in order to score well. A dozen or so questions always relate to geometry concepts. While trigonometry is not an official topic for the SAT, you do need to know how and when to apply the Pythagorean Theorem for certain problems. Don’t worry though: You don’t need to memorize the formula because a formula table is provided at the beginning of every single SAT math section.

We want to emphasize that the SAT won’t just be about plugging numbers into the formula and chugging out an answer; it will require you to use reasoning skills to determine when it’s appropriate and useful to apply the Pythagorean Theorem. In addition, there will be times when it won’t even be obvious that you can apply the formula in order to help you solve a problem. You will then need to look for “hidden triangles” in the problem to help you reach the solution. Once you know what to look for, you’ll start seeing hidden right triangles everywhere and be able to use them to your advantage to solve difficult geometry SAT problems.

What is the perimeter of the trapezoid above?

A) 52

B) 72

C) 75

D) 80

E) 87

Do you remember the formula for finding the perimeter of a trapezoid in this situation? You don’t? That’s all right because there isn’t one. Then how should we go about solving this problem? This is where the “hidden triangle” concept comes into play. The problem here is that we don’t know the length of the base of the trapezoid, so we’ll need to get a little creative to solve for it. Taking a closer look, we could draw a straight line from the top of the trapezoid to the base to form a triangle as follows:

There’s our hidden triangle. Now we have a rectangle with base 20 and sides 15 attached to a right triangle. If we can figure out the base of the triangle on the left, we’ll have all the components needed to figure out the perimeter of the entire shape. Let’s go ahead and apply our Pythagorean Theorem where 17 is the hypotenuse and 15 is one of the sides, and we’ll let x be the remaining side:

172 = 152 + x2

289 = 225 + x2

64 = x2

x = 8

Now that we have the base of the triangle, we know that the base of the trapezoid is simply 20 + 8 = 28. Let’s add the other sides together for: 28 + 17 + 15 + 20 = 80, or answer choice D.

You see how hidden triangles have helped us to solve this geometry problem? Look for them the next time you come across an SAT geometry problem.

If you were completely stumped on this problem, you could have also used your SAT test prep strategies to eliminate some incorrect answer choices. We know that the first three sides add up to 52 (17 + 20 + 15). We also know that the base of the trapezoid must be greater than the top or greater than 20. As a result, we know that the perimeter is greater than 72. This allows us to eliminate some answer choices that are too small:

A) 52

B) 72

C) 75

D) 80

E) 87

On top of that, you could probably safely eliminate answer choice C for being too small a value and answer choice E for being too large a value, based on the relative lengths of the rest of the figure. In any case, you gain an edge on this problem and would be in good guessing territory even if unable to reach the full solution. Some clever elimination strategies can be worth a handful of points on the SAT.

Original Source:http://www.businessweek.com/articles/2014-02-19/sat-geometry-crouching-trapezoid-hidden-triangle

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